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\author{五六七 }
\title{带空气阻力的单摆运动 }

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\maketitle

\begin{abstract}
建立单摆运动的微分方程，考虑空气阻力。并用观测数据进行验证。
\end{abstract}

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\section{问题描述}
一根长度为 $L$ 的细绳，挂一个质量为 $m$ 的重物，以初始角度 $\theta$ 从静止开始摆动。求摆动的角度与时间的函数表达式。考虑空气阻力，摆动多久会静止？

\begin{figure}[ht!]\centering
\includegraphics [height=4cm, width=7cm]{pendulum.png}
\caption{一个单摆的图示 }
\end{figure}

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\section{建立模型}
\subsection{不考虑空气阻力}
摆动的路径是一段圆弧。重力在切线方向上的分量，根据牛顿第二定律，就是重物的加速度。
\begin{figure}[ht!]\centering
\includegraphics [height=4cm, width=7cm]{pendulum-2.png}
\caption{重力的分量与摆动角度的关系 }
\end{figure}

设摆动的角度与时间的函数关系是 $\theta=\theta(t)$, 则速度大小为 $L\theta'(t)$, 加速度大小为 $L\theta''(t)$. 
由此可得微分方程
\begin{eqnarray}
mg\sin\theta(t) = mL\theta''(t).
\label{eqn-1}
\end{eqnarray}
这是二阶微分方程，写成另一种形式，可得
\begin{eqnarray}
\frac{d^2\theta}{dt^2} = a^2\sin\theta, 
\label{eqn-2}
\end{eqnarray}
其中 $a^2=\frac{g}{L}$ 是一个正常数。

当摆动角度 $\theta$ 很小时，$\sin\theta$ 近似于 $\theta$. 这时可将微分方程近似为线性常微分方程
\begin{eqnarray}
\frac{d^2\theta}{dt^2} = a^2\theta 
\label{eqn-3}
\end{eqnarray}
可得通解为 
\begin{eqnarray}
\theta = c_1\cos(at)+c_2\sin(at), 
\label{eqn-4}
\end{eqnarray}
其中 $c_1,c_2$ 是任意常数，可以根据初始位置和初始速度确定。

微分方程(\ref{eqn-2})的解析解，不是一个初等函数。具体表达式跟一个椭圆积分有关。


\subsection{考虑空气阻力}

现在考虑空气阻力。设阻力大小与速度的函数关系为 $f=f(v)$, 则微分方程(\ref{eqn-1})变为
\begin{eqnarray}
mg\sin\theta(t) - f(L\theta'(t)) = mL\theta''(t).
\label{eqn-5}
\end{eqnarray}

分别考虑 $f(v)=-kv$, $f(v)=-k\sqrt{v}$ 与 $f(v)=-kv^2$ 进行讨论，其中 $k$ 是待定参数。拟合观测数据，分析哪个模型更符合实际情况。也可以考虑其它阻力模型。


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\section{编程计算}





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\section{回答问题}




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%\section{参考文献 }
\begin{thebibliography}{99}
\bibitem{dingtongren} 丁同仁、李承治，\emph{常微分方程教程}，高等教育出版社，2022年3月第三版。
\bibitem{sishoukui-2} 司守奎,孙玺菁. \emph{Python数学建模算法与应用}, 国防工业出版社. 2022年1月第1版. 


\end{thebibliography}

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